Lemma: Local Boundedness
If \(\lim\limits_{x \to a} f(x) = L\), then there exists an \(M > 0\) and \(\delta > 0\) such that \(f(x) \leq M\) for all \(|x - a| < \delta\).
Proof
Let \(\epsilon = 1\). Then by defintion there exists a \(\delta > 0\) such that \(|x - a| < \delta\) implies that
$$
\begin{align*}
|f(x)| - L \leq |f(x) - L| < 1
\end{align*}
$$
Hence, \(|f(x)| \leq 1 + L\) if \(|x - a| < \delta\). \(\ \blacksquare\)
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai